Problem: Evaluate $\int\dfrac{\cos^2x}{1-\sin x}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $x+\cos x+C$ (Choice B) B $x-\cos x+C$ (Choice C) C $x-\sin x+C$ (Choice D) D $x+\sin x+C$
Answer: First use the identity $~\cos^2x = 1-\sin^2x~$ to rewrite the integral. $ \int\dfrac{\cos^2x}{1-\sin x}\,dx=\int\dfrac{1-\sin^2x}{1-\sin x}\,dx$ Now factor the numerator as the difference of two squares and simplify. $ \begin{aligned}\int\dfrac{\cos^2x}{1-\sin x}\,dx&=\int\dfrac{(1-\sin x)(1+\sin x)}{1-\sin x}\,dx\\ \\ \\&=\int(1+\sin x)\,dx\end{aligned}$ Now the integration is straightforward. $ \begin{aligned}\int\dfrac{\cos^2x}{1-\sin x}\,dx&=\int(1+\sin x)\,dx\\ \\ \\&=x-\cos x+C\end{aligned}$